Find the EMF of the source in the circuit The following is the question:

For driving a current of $2 A$ for $6  $ Minutes in a circuit, $1000J$ of work is to be done.   Find the EMF of the source in the circuit.

$My$ $Attempt:$
The amount of charge flowed in $6$ minutes is $2A\times360s = 720 C$, the $V_T = \frac{1000}{720} = 1.38 V$, where $V_T$ is the terminal resistance. Now, $\varepsilon = I \cdot R_{IN} + V_T $ , where $R_{IN}$ is the internal resistance of the source. So, $\varepsilon = (2A )\cdot R_{IN} + 1.38$.
How to proceed from here ?
The official answer is $1.38$, but I am confused by it because its equal to $V_T$.
 A: The question doesn't tell you WHERE the work is to be done. To get the answer given for the emf, you have to assume that the 1000 J is the total work done in the circuit, that is the sum of the work done inside the cell $(I^2R_{IN}\times 360\  \text s$ with your notation) and the external work done $(IV_T \times 360\ \text s$).
By definition, the emf, $\mathscr E$, is the total work done in the circuit per unit charge flowing. Therefore if 1000 J is the total work,
$$\mathscr E \times 2\ \text A \times 360\ \text s = 1000\ \text J$$
So $\mathscr E =1.39\ \text {J C}^{-1}=1.39\ \text V$
By writing $V_T = \tfrac{1000}{720}= 1.39\ \text V$ you were assuming that the 1000 J was used in the circuit external to the battery. This assumption would also give you $\mathscr E = 1.39\ \text V$ if you also assume that the internal resistance, $R_{IN}$, is zero. In this case there is no work done by charge moving through the cell itself; in other words the total work is equal to the work done outside the cell!
